Two circular coils P and Q of 100 turns each | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 N I}{2r}$.Given: $N = 100$,$r = \pi 	ext{ cm} = \pi 	imes 10^{-

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 N I}{2r}$.Given: $N = 100$,$r = \pi 	ext{ cm} = \pi 	imes 10^{-2} 	ext{ m}$.For coil $P$ with $I_1 = 1 	ext{ A}$:$B_P = \frac{\mu_0 	imes 100 	imes 1}{2 	imes \pi 	imes 10^{-2}} = \frac{4\pi 	imes 10^{-7} 	imes 100}{2 \pi 	imes 10^{-2}} = 2 	imes 10^{-3} 	ext{ T} = 2 	ext{ mT}$.For coil $Q$ with $I_2 = 2 	ext{ A}$:$B_Q = \frac{\mu_0 	imes 100 	imes 2}{2 	imes \pi 	imes 10^{-2}} = \frac{4\pi 	imes 10^{-7} 	imes 200}{2 \pi 	imes 10^{-2}} = 4 	imes 10^{-3} 	ext{ T} = 4 	ext{ mT}$.Since the planes are mutually perpendicular,the magnetic fields $B_P$ and $B_Q$ are perpendicular to each other.The resultant magnetic field is $B_{	ext{net}} = \sqrt{B_P^2 + B_Q^2} = \sqrt{(2)^2 + (4)^2} = \sqrt{4 + 16} = \sqrt{20} 	ext{ mT}$.Comparing with $\sqrt{x} 	ext{ mT}$,we get $x = 20$.

Similar Questions

The magnetic field at the origin due to the current $I$ flowing in the wire as shown in the figure is:

An element $\overrightarrow{\Delta \ell} = \Delta x \hat{i}$ is placed at the origin and carries a current of $10 \ A$. Find the magnitude of the magnetic field on the $Y$-axis at a distance of $0.5 \ m$,given $\Delta x = 1 \ cm$. (Use $\frac{\mu_0}{4 \pi} = 10^{-7} \ T \cdot m/A$)

Explain the experiment which demonstrates the magnetic field produced due to a straight long current-carrying wire.

$A$ long straight wire, carrying current $I,$ is bent at its midpoint to form an angle of $45^{\circ}.$ The magnetic field induction at point $P,$ at a distance $R$ from the point of bending, is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module